Nuclear Fuel Cycle

Problem: Calculate the specific activity (in Bq per gram) of UO$_2$ fuel enriched to 3.5 wt% U-235.

The formula for specific activity is:

$A = \lambda N = \frac{\ln 2}{T_{1/2}} \times \frac{m \times N_A}{M}$

Where:

• $A$ = activity (Bq, i.e. disintegrations per second)
• $\lambda$ = decay constant (s$^{-1}$) = ln(2) / $T_{1/2}$
• $N$ = number of atoms
• $T_{1/2}$ = half-life (in seconds)
• $m$ = mass of the isotope (g)
• $N_A$ = Avogadro’s number = 6.022 $\times$ 10$^{23}$ mol$^{-1}$
• $M$ = atomic mass of the isotope (g/mol)

Useful conversion: 1 year = 3.156 $\times$ 10$^7$ seconds

Step 1: Determine the mass fraction of uranium in UO$_2$

The molecular mass of UO$_2$ is:

$M_{\text{UO}_2} = 238 + 2 \times 16 = 238 + 32 = 270 \text{ g/mol}$

Therefore, the mass fraction of uranium in UO$_2$ is:

$f_U = \frac{238}{270} = 0.8815$

This means that in 1 gram of UO$_2$, there is 0.8815 g of uranium.

Tip: A very common student error is to forget this step and assume that 1 g of UO$_2$ contains 1 g of uranium. Always account for the oxygen content!

Step 2: Determine the isotopic composition of 3.5% enriched uranium

For enriched uranium at 3.5 wt% U-235, the enrichment process also increases the proportion of U-234. A reasonable approximation for 3.5% enriched fuel is:

• U-234: Weight fraction 0.0267%; mass in 1 g UO$_2$ = 0.8815 $\times$ 0.000267 = 2.35 $\times$ 10$^{-4}$ g
• U-235: Weight fraction 3.50%; mass in 1 g UO$_2$ = 0.8815 $\times$ 0.035 = 3.085 $\times$ 10$^{-2}$ g
• U-238: Weight fraction 96.47%; mass in 1 g UO$_2$ = 0.8815 $\times$ 0.9647 = 8.504 $\times$ 10$^{-1}$ g

Note: When uranium is enriched in U-235, the U-234 proportion also increases (because U-234 is lighter than U-235, so it concentrates preferentially in centrifuge enrichment). For 3.5% enrichment, U-234 increases from 0.0055% to approximately 0.027%.

Step 3: Calculate the activity of each isotope (per gram of UO$_2$)

U-234:

$T_{1/2} = 2.46 \times 10^5 \text{ years} = 2.46 \times 10^5 \times 3.156 \times 10^7 = 7.76 \times 10^{12} \text{ s}$

$\lambda_{234} = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{7.76 \times 10^{12}} = 8.93 \times 10^{-14} \text{ s}^{-1}$

$N_{234} = \frac{m \times N_A}{M} = \frac{2.35 \times 10^{-4} \times 6.022 \times 10^{23}}{234} = 6.05 \times 10^{17} \text{ atoms}$

$A_{234} = \lambda_{234} \times N_{234} = 8.93 \times 10^{-14} \times 6.05 \times 10^{17} = 5.40 \times 10^{4} \text{ Bq}$

U-235:

$T_{1/2} = 7.04 \times 10^8 \text{ years} = 7.04 \times 10^8 \times 3.156 \times 10^7 = 2.22 \times 10^{16} \text{ s}$

$\lambda_{235} = \frac{0.6931}{2.22 \times 10^{16}} = 3.12 \times 10^{-17} \text{ s}^{-1}$

$N_{235} = \frac{3.085 \times 10^{-2} \times 6.022 \times 10^{23}}{235} = 7.90 \times 10^{19} \text{ atoms}$

$A_{235} = 3.12 \times 10^{-17} \times 7.90 \times 10^{19} = 2.46 \times 10^{3} \text{ Bq}$

U-238:

$T_{1/2} = 4.47 \times 10^9 \text{ years} = 4.47 \times 10^9 \times 3.156 \times 10^7 = 1.41 \times 10^{17} \text{ s}$

$\lambda_{238} = \frac{0.6931}{1.41 \times 10^{17}} = 4.92 \times 10^{-18} \text{ s}^{-1}$

$N_{238} = \frac{8.504 \times 10^{-1} \times 6.022 \times 10^{23}}{238} = 2.15 \times 10^{21} \text{ atoms}$

$A_{238} = 4.92 \times 10^{-18} \times 2.15 \times 10^{21} = 1.06 \times 10^{4} \text{ Bq}$

Step 4: Calculate total specific activity

$A_{\text{total}} = A_{234} + A_{235} + A_{238}$

$A_{\text{total}} = 5.40 \times 10^4 + 2.46 \times 10^3 + 1.06 \times 10^4$

$\boxed{A_{\text{total}} \approx 6.7 \times 10^4 \text{ Bq per gram of UO}_2}$

Or approximately 67 kBq/g.

Summary of Contributions

Key insight: Even though U-234 makes up only 0.027% of the uranium by mass, it contributes approximately 81% of the total activity. This is because its half-life is about 2,800 times shorter than U-235 and about 18,000 times shorter than U-238.

Tip: In examination questions, if you are given natural uranium (not enriched), you will use the natural isotopic fractions (U-234: 0.0055%, U-235: 0.72%, U-238: 99.27%). The calculation method is identical.